 Hi, I'm Zor. Welcome to a new Zor education. I will talk about limits. Probably, at least in the beginning, I would talk only about limits of the sequences. So we are talking about sequences and their limits. Now, obviously we are talking about infinite sequences, which means we have elements of the sequence number one, number two, number three, number n, number n plus one, and any other n up to infinity. So for any n, we have defined an element of the sequence, defined for any n, any integer positive n. So that's number one. Well, number two, let's just think about defining what actual limit is. Now, intuitively, we all understand what does it mean to tend to something or to converge to something. If I will ask you what is the limit of this sequence, et cetera, when n is infinitely increasing, I'm sure everyone would answer yes, the limit is zero basically because the numbers are getting smaller and smaller and smaller. Well, I agree. However, that's not the definition. We really have to define it and distinguish this, for instance, from, I don't know, let's say one tenths plus one half, one tenths plus one third, one tenths plus one tenths, et cetera. These are also diminishing, but they are diminishing to one tenths, not to zero. So somehow we have to define what it means to get closer and closer to some particular number. And that's what I would attempt to do. So instead of just giving you the definition, let's try to just think about how to define it properly. All right, so first of all, we are talking about some number L, which is supposed to be a proposed limit for this particular sequence. Now we are talking about converging of that particular sequence to this limit, which means basically getting closer and closer. So number one, does it mean that every sequential element of this particular sequence goes closer and closer to L? Well, quite frankly, not necessarily, because what you might actually imagine is, let me do it graphically. Yes, this is a sequence which tends to zero, but so is this sequence. It also tends to zero, but increasing its numbers. So it's basically rising to its limit. And I can even give you this particular set of elements. It also tends to zero because it's getting closer and closer to the zero level. But it goes up and down, up and down, up and down. So sometimes it's going to this particular limit, then it goes outside of it, then again towards and then inside, increasing and decreasing, et cetera. So we have to basically cover all these complicated cases in our definition of the limit. And again, the closeness is not monotonal, monotonous in this particular case. Some elements are on the limit already, but then some subsequent element can go out of it, although the distance from the limit is actually decreasing. Now, how can we specify it? So my first suggestion is the following. Number one, we have to say that we are going to the limit closer and closer in the following term. For each distance, d, greater than zero, from the limit, we will have, we will find elements at certain number n greater than capital N. So we have to find some kind of a capital N number after which the elements of our sequence would be closer to the limit than this particular distance, d. So for any instance, d, I have to find the number after which elements will be closer than this distance. So in this particular case, let's have this distance. This is my d and this is my minus d. So this is the corridor and as you see, eventually after something like this, this is n, all members of my sequence are within this corridor. And what's important is that for any positive distance, d, from the limit, I should find the number n after which my sequence is within this corridor around the limit. And d can be chosen any small number, like 1 tenths. Well, then we will find this particular number. If I will ask for 100s, for instance, then the number will be further, but we will still be able to find the number. So let me just summarize it again. For any distance, d from the limit, from the proposed limit, we have to find the number n, capital N, after which all subsequent members elements of our sequence will be within this particular distance from the limit. Only if this is satisfied, we can say that our sequence has a limit. So again, l is a limit for a sequence a n, when and only when for any positive d, however small, we can find the sequence number n after which all members will be within the distance d from this limit. That's what it means that the particular sequence has a limit called l. Sometimes we're using the word convergent, I already used it a couple of times. So if l is the limit of the sequence a n, we say that a n converges to l. So that's as much about the definition. Now, do all the sequences have limits? Well, definitely not. Let's say one minus one, one minus one. This sequence doesn't have a limit. It goes up and down, up and down. There is no single number to which it converges. Well, there are only two candidates actually, one and minus one, but none of them is actually limit because one cannot be a limit because let's say on the distance of one third from one, I will find infinite number of minus ones which are not in that particular corridor around one. And same thing about minus one. So this is not a convergent sequence. There are sequences which do not have limit. Something just came to my mind. Can the sequence have two limits? Well, think about it. Obviously not because if the sequence is within certain corridor around one particular limit so the values are supposed to be eventually here. At the same time, if we have another so-called limit, it should be within a very small distance around this particular limit after certain number m, right? So if we will choose these corridors in such a way that they're not overlapping each other, then you'll see that this is impossible because if after certain n one, it's in this corridor and under certain n two, it's in this particular corridor, these two are impossible to satisfy at the same time for any n which is greater than maximum of these two, right? So basically it's impossible to have two different limits. But this is an obvious kind of stuff. There are a little bit less obvious properties of limits which I'm going to address. I actually put it here. So number one, a convergent sequence is bounded which means there is some number which is greater than any element of this sequence and there is another number which is less than any element of this sequence. How can it be proven? Well, very easily. So if you have a n and you have a limit, let's designate it as this arrow. So it has a limit. Now it means that for let's just fix some kind of a d greater than zero. Doesn't really matter what exactly this value is. But we will find a number n such that if smaller n is greater than the capital N, the distance between these values for any n greater than capital N. It means that the tail of this particular sequence is already limited. It's no more from this we can actually derive that it's less than ld, l plus d and greater than l minus d. So the tail of a n sequence, all the elements with order numbers greater than this particular capital N which we found for this d. All the tail is limited. We have it on the bottom and on the top bounded by these two values. Now how about the previous from one to N? Well, obviously there are finite number of these elements. There are only n elements before this particular property gets into effect, right? So since it's a finite, it has minimum and maximum. And they are also finite. These are numbers, a one, a two, et cetera, a n. Now after a n, they're bounded by these guys. Before a n, there are only finite number of these guys which means we can always choose minimum and maximum. Well, if this minimum, for instance, is less than l minus d, then this minimum actually bounds the whole sequence from below. If this minimum is not less than l minus d, then l minus d is basically bounding the whole sequence. So that's about the bottom. How about the top? Same thing. Take the maximum from these elements from a one to a n. Well, if this maximum is greater than l plus d, then the maximum will be the bound, the upper bound for the whole sequence. If it's less than l plus d, then l plus d is the bound, the upper bound for the whole sequence. But in any case, considering the infinite tail is bounded by these guys, and we have only finite number of elements before that, we can always find upper and lower bound. So a sequence which converges to some limit is always bounded by some minimum and maximum value. Sequence which does not have a limit might or might not be bounded, by the way. The sequence one minus one, one minus one, one minus one, one minus one, et cetera, doesn't have a limit, but it's bounded. It's from minus one to one. But the sequence, let's say one, two, three, four, five, six, et cetera, it doesn't really converge to anything, and it's unbounded on the top. A sequence zero, one, minus one, two, minus two, minus two, three, minus three. It also has no limit, and it's unbounded on both sides. It can be greater than any positive number, and it can be less than any negative number, so it's unbounded on both sides, upper and lower. But if it's a convergent sequence, it's bounded. Now, we will use this property at some point in the future, but let's go to other properties. Now, these other properties seems to be a little bit more obvious, but they still need to be proven, and it's the proof which actually, the purpose of this explanation. It's obvious that they are true, intuitively obvious, but the proof is always interesting because it has theoretical kind of value, and it's a good exercise for the brain. All right, property number one. If I have a sequence a n, which tends to limit l, and then I have some multiplier, k, I claim that k times a n, which is a new sequence, converges to k l. It's kind of obvious because we are increasing or decreasing whatever every member of our sequence at some factor, at some rate, and obviously the limit which this particular sequence converges to should also be multiplied by the same factor. But how can I prove this? Well, let's just go from the definition of the limit. Let's remember again, for any d greater than zero should exist n such that if n greater than zero then l minus a n is within the d. Okay, now in this particular case, this is our sequence. Now for any d we have to find n such as in this particular case k l minus k a n should be less than d. Now, how can I achieve this? Very simply, now I choose d, right? Now I know that a n converges to l. So I will take d over k as a new distance, but I will apply it to this particular thing. Well, let's just postulate that k is not equal to zero because for k equals to zero it's obvious because these are all zeros and this is zero, so it's okay. Right, so k is not equal to zero. Now I choose some kind of a d greater than zero and I have to find a corresponding number and when this will be actually true. But now what I will do, I will choose d over k and I will apply the convergence of a n to l. Now what does it mean? There is some kind of a n for this particular distance after which for n greater than n, a n is at the distance d over k. Right, since a n converges to l, then for any d including this d over k, I can find my number n when this equation, when this inequality is true. But this inequality is exactly equivalent to this one as you see. You just multiply k and put it inside the absolute value thing. So basically for any d which I choose to determine convergence of this thing, I use a convergence of the original sequence for d over k and find the number n when this is true. And since this is true, then for the same number n, this will also be true because I can multiply both sides by absolute value of k. That's the proof. We have found the n when this is always true for any lower case n greater than capital N. We found this n. Since we found it, it means that for any d, we found the number when this particular inequality holds for any subsequent number. That proves the convergence of this sequence to this one. So in short, we can say that the limit of multiple of sequence is equal to multiple limit. What's the word? Limit of multiple of a sequence is equal to multiple of limit. Right, more or less. Okay, now, analogous. Limit of sum of sequence is equal to sum of limits, which means if I have two different sequences, one converges to a and another to b. Then a new sequence, which is their sum, the corresponding sum, the first and first, second and second, etc. So this new sequence converges to the sum of limits. So again, limit of the sum is equal to sum of the limits. How can that be proven? Again, let's do it from the definition. That's the best way. Start from the definition. By the way, whenever you have an argument with anybody, like one person says one thing and another person says another thing, start from the definition. Make sure you're talking about the same thing. Because one person is saying, okay, democracy is good and another person says democracy is bad. Well, before discussing this, why don't you define democracy and make sure that you have exactly the same definition of democracy? So that's very important. So let's start from the definition. Okay, now, definition means we have to prove that for any d greater than 0, we can find n such as a plus b minus a m plus b m, absolute value less than g for lowercase n greater than capital N. So that's what we have to do. For any d, we have to find a number n capital after which this would be true for any n greater than capital N. Well, it's again not very difficult thing to do, because obviously I can regroup it this way. a minus a m plus b minus b m. Now, let me point out some property. Property of absolute value. I hope everybody knows that this is true. The absolute value of sum is always no greater than sum of absolute values. Well, for both positive x and y, that's an equality of this, because every member is positive. When one of them is negative, then let's say x is positive and y is negative. Then you're basically reducing x, and that's why we have a straight less than sign. If x is negative and y is positive, same thing. And if both are negative, then again we will have an equality in this particular case. So this is kind of an obvious property of absolute value, which can be proven a little bit more rigorously. I just gave you the idea of the proof, which is kind of obvious. But I will use it here. You see, this is less than or equal than a minus a m plus b minus b m, right? x in this case is a minus a m, and y is b minus b m. Now, how can I use this to find capital N for capital N for d greater than 0? This is not good anymore. So for any d, I have to find m when this thing is true. But since this is greater, if I will find the n when this is true, I will solve my problem, right? If I will find the capital N when this is true, since this is less, it will be also true. This will also be less than d. Now, how can I find this n so this is true? Well, very simply, since a n converges to a, I can find m1 such that capital A minus a n is less than d over 2. Now, since b n converges to b, I can find another number n2 for d2 also when the difference is also less than d2. Now, if I will have n equal to maximum of m1 and m2, both inequalities will be felt true, right? Because this is for any lowercase n greater than n1. And this is for any lowercase n greater than n2. So for any lowercase n greater than maximum between these two numbers, I will have both felt true. And if I will summarize them together, I will get this and now I will use this inequality to state that this also will be less than g. So that's how we find our number n for which some of these two sequences would be within the distance d from the sum of the limits. First, I find m1 when the first sequence is within the distance d over 2 from the a. Then I find another n2 when the second sequence is within the distance d2 from its limit and maximum among these two numbers would give me the number after which both sequences will be within the limit d over 2 from their corresponding limits and therefore their sum would be within the distance d from sum of the limits. So that's about sum of these two sequences. How about product? Product is also has exactly the same property. And two sequences have corresponding limits. Their product converges to the product of the limits. Again, the proof. Exactly the same thing. Choose any distance d and let's find the number after which a b minus i n d n is less than g. How can I find the number n after which all the members have this property? Okay. Let me just transform this slightly. If you remember when I had addition, sum of two sequences, transformation was basically regrouping of numbers. So a plus b minus a n plus b n, I just regrouped a minus a n and b minus b n. In this case, it's not as easy. But there is a trick. a b minus a b n plus a b n minus a n b n. That's what they did. So I subtracted a times b n and added a times b n. And this is exactly the same thing since they cancel each other, right? But now I will do this. I'll group this one and this one. Now, this one would be a times b minus b n. And this one would be a minus a n times b n less than g. Now I will use exactly the same property of the absolute value that the absolute value of sum of these two is less or equal than absolute value... Actually, this is equal. I'm just regrouping this. And now I will break it into two sums of absolute values. Absolute value of a times b minus b n plus absolute value of a minus a n times b n. So this expression is less than this. So what I will do, I will find my capital N after which this would be less than g. How can I do that? Let me simplify it even more. So this thing is equal to absolute value of a times absolute value of b minus b n plus absolute value of a minus a n times absolute value of b n. Now, using the fact that a n converges to a and b n converges to b, given my distance d, I can actually choose numbers after which this will be sufficiently close to zero and this will be sufficiently close to zero. So the whole thing would be less than d. Now, which numbers should I choose? Well, in this case it's easy. I choose N1 such that b minus b n is less than d over absolute value of a. If absolute value of a is equal to zero in the first place, I don't even have to do this. I immediately just cancel the whole thing and say, okay, it's zero anyway. It's a separate case obviously, but it's a very trivial case. So I don't have any doubts by dividing by a. Now, in this case, it's a little bit more complex. But now let's remember the first property of the sequence limits of the sequence. If sequence converges to certain limit, then it's bounded basically. There is a minimum and there is a maximum. So the absolute value of the sequence is obviously bounded as well by zero on one side and whatever the maximum is on another side. So basically what I will do is I will do the maximum of b n of absolute value of b n, which is a positive case. Now, this thing, let's say, I can always say that absolute value of b n is less than some kind of a constant c. Since there is a convergence, there is always this property. There is some kind of a maximum of absolute value of b n about which it doesn't really go. So here I can choose n2 such as a minus a n is less than g over c, this c. Now, if I choose this n1 using the property of b n to converge to b, so for all n greater than n1, this is true. And I will choose capital N2 so that for all n greater than n2, this thing is true. Then, again, for n equals maximum of n1 and n2, for any n, I have both my inequalities held for any lowercase n greater than this uppercase n. And since both inequalities are true, then I can obviously multiply this by a. I will have this is less than, let's say, actually, I think I would better actually have this d over 2. It doesn't really matter. Since it converges, now I will have exactly the d. So if I find the capital N1, so this is less than d over 2, absolute value of f a. And here it would be less than d over 2 c. Then my this expression would be less than, this would be multiplied by 2a by a and it would be less than d over 2. So this is less than d over 2 plus, and this expression, since this is this way, now it would be, instead of this, I will put this. Now, absolute value of bn over c would be obviously less than 1, which means the whole thing will also be less than 1. Okay, so that's d over 2 plus d over 2, which is d, and that's exactly what we wanted. So how can I find the number N, after which the original difference between a, b, and a and bn would be less than d? Well, first I find this n1 from the convergence of bn, then I find this n2 from the convergence of an to a, where c is upper boundary for bn, perhaps lower boundary for bn, and then the maximum of these two would be the number N, after which I can say that this would be closer to this than the distance d. Okay, and the last property which I wanted to address is inverse. Well, as you, I'm sure, guessed, it's also true that if you inverse a sequence, then if the sequence tends to a certain limit, then the limit should be actually inverse as well with certain restrictions of this. So if a n tends to l, then 1 over a n tends to 1 over l. The only thing is, obviously, it's true for both all elements of the sequence and its limit not equal to 0. So let's consider the sequences which do not have elements equal to 0 and which, the limits of which is not equal to 0, neither. So for these cases, this is true. Now, how can I prove it? Just think about 1 over l minus 1 over a n, absolute value, is a n minus a n minus l divided by l a n. Now, a n as a convergent sequence is bounded. Again, we are using the property of the limited sequence to be bound. Now, since it's bound, how can I find the number n? So this is war equivalently a n minus l less than d times l times, let's call it c, where c is the maximum or the minimum. Let me just think about it. If c is maximum, then this would be this way. Now, c should be the minimum. Okay, so this is a minimum of absolute value of a n. Now, the a n is a bounded series which does not have the values of 0 among them and limit is not equal to 0. Therefore, the bound is not equal to 0, obviously. So if I take the minimum and I will find the a n, I will find number n after which the original sequence has a distance from its limit less than this for any given d. So first I get d and from d I'm getting this particular inequality. I find the n. So if this is true, then obviously this which is less than this and this is less than d. I'm still kind of confused whether it's maximum or minimum. It should be c. I mean it works both ways actually, but let me think. So this should be less than this, which means this should be greater than this. So if denominator is greater, then the fraction is smaller. Yeah, I'm right. That's the minimum. We divide by a smaller number, and that's why we have a bigger fraction. So again, for any d we find the n when this particular inequality is true and from this follows this and this. So again, for any d we found number n after which the difference between the elements of the sequence and limit are smaller than that d, whichever, however small we choose this particular d. So let me just summarize the whole thing. Basic operations with sequences which have limits, which are convergent to certain limits. Basic operations like addition, multiplication by some number, multiplication of two sequences, or inversing the sequence when nothing is equal to zero. These particular operations are working with sequences in exactly the same fashion as with their limits, like some of the sequences which have limits is equal to, tends to some of the limits. That's basically the properties of the limits which I wanted to explain. There will be a certain number of problems obviously in the future lectures about finding the limits and proving that these are limits. Like I try to prove it here for individual properties, but that will be in the future lectures now. For today, that's it. Try to go through these proofs again just by yourself. They are in notes on the website unisor.com. Please listen to the problems. The problems are very, very important. I do encourage you to try to solve them yourself first, and then go to the corresponding lecture, see how I suggest the solution or a proof or whatever. But then try to gain by yourself as well, so it will imprint in your mind. That's it for today. Thank you very much.